Converting InFix to PostFix with Linked Stacks

[mlanuri10]

#include <iostream> // <<, >>, cout, cin
#include <string> // string, ==, find, npos
#include <cassert> // assert()
#include <cctype> // alnum()
using namespace std;

#include "LStack.h"

string PostFixExp(string exp);

int main()
{
string InFixExp;
cout << "Type 'X' to leave the program." << endl;

for (;;)
{
cout << "Enter an expression: ";
getline(cin, InFixExp);

if (InFixExp == "X") break;

cout << "Postfix expression is: " << PostFixExp(InFixExp) << endl;
}
exit(0);
}

string PostFixExp(string exp)
{
char * token,
topToken;
Stack opStack;
string PostFix;
const string BLANK = " ";

for (int i = 0; i < exp.length(); i++)
{
token = &exp[i];
switch (*token)
{
case ' ' : break; //if it's space, continue
case '(':
opStack.push(*token); //if it's left parenthesis, push on the stack.
break;
case ')':
for (;;) //if it's righ parenthesis, pop it.
{
assert(!opStack.empty());
topToken = opStack.top();
opStack.pop();
if (topToken == '(') break;
PostFix.append(BLANK + topToken);
}
break;
case '+': case '-': //check for any operations
case '*': case '/': case'%':
for (;;)
{
if (opStack.empty() ||
opStack.top() == '(' ||
(*token == '*' || *token == '/' || *token == '%') &&
(opStack.top() == '+' || opStack.top() == '-'))
{
opStack.push(*token);
break;
}
else
{
topToken = opStack.top();
opStack.pop();
PostFix.append(BLANK + topToken);
}
}
break;
default: // operand
PostFix.append(BLANK + token);
for (;;)
{
if (!isalnum(exp[i + 1])) break; // end of identifier
i++;
token = &exp[i];
PostFix.append(1, *token);
}
}
// Pop remaining operators on the stack
for (;;)
{
if (opStack.empty()) break;
topToken = opStack.top();
opStack.pop();
if (topToken != '(')
{
PostFix.append(BLANK + topToken);
}
else
{
cout << " *** Error in infix expression ***\n";
break;
}
}
}
return PostFix;
}

What can I fix this order in order to work properly?

[kbw]

1. Please use the code formatting tags to format your code. It's not readable at present.
2. Why do you think there's something wrong with it?